Problem 33 - Thomas Calculus

Question

Converting to a polar integral Integrate \(f(x, y)=\) \(\left[\ln \left(x^{2}+y^{2}\right)\right] / \sqrt{x^{2}+y^{2}}\) over the region \(1 \leq x^{2}+y^{2} \leq e\)

Step-by-Step Solution

Answer

The integral evaluates to \(4\pi - 2\pi\sqrt{e}\).

Step 1: Understanding Polar Coordinates

In polar coordinates, every point in the plane is determined by an angle \(\theta\) and a radius \(r\). The relationships are:\(x = r\cos(\theta)\) and \(y = r\sin(\theta)\). Thus, \(x^2 + y^2 = r^2\).

Step 2: Set up the Integral in Polar Form

Convert the integral from Cartesian (\(x, y\)) to polar (\(r, \theta\)) coordinates. The given region is \(1 \leq x^2 + y^2 \leq e\), which converts to \(1 \leq r^2 \leq e\) or \(1 \leq r \leq \sqrt{e}\). Also, \(f(x, y)\) becomes \(\frac{\ln(r^2)}{r}\) in polar coordinates.

Step 3: Adjusting the Differential

In polar coordinates, the area element \(dx\,dy\) becomes \(r\,dr\,d\theta\). Hence, the integrand \(\frac{\ln(r^2)}{r}\) is adjusted to \(\ln(r^2)\,dr\,d\theta\).

Step 4: Set Up the Double Integral

The region of integration is a circular annulus. The limits for \(r\) are from 1 to \(\sqrt{e}\), and for \(\theta\), it's 0 to \(2\pi\). The integral in polar coordinates is: \[ \int_{0}^{2\pi} \int_{1}^{\sqrt{e}} \ln(r^2) \, dr \, d\theta \]

Step 5: Simplifying the Integrand

Notice \(\ln(r^2) = 2\ln(r)\). Replace this in the integrand to make it easier to integrate: \[ \int_{0}^{2\pi} \int_{1}^{\sqrt{e}} 2\ln(r) \, dr \, d\theta \]

Step 6: Integrate with Respect to r

Perform the inner integration over \(r\): \[ \int_{1}^{\sqrt{e}} 2\ln(r) \, dr \] To solve this, let \(u = \ln(r)\) making \(du = \frac{1}{r} dr\). Then the integral becomes \[ 2 \int \ln(r) \, dr = 2 \left[ r\ln(r) - r \right]_{1}^{\sqrt{e}} \]

Step 7: Evaluate the Integral

Calculate: \[ 2 \left[ \sqrt{e}\ln(\sqrt{e}) - \sqrt{e} - (1\ln(1) - 1) \right] = 2\left[ \sqrt{e} \cdot \frac{1}{2} - \sqrt{e} + 1 \right] \]This simplifies to:\[ \sqrt{e} - 2\sqrt{e} + 2 = 2 - \sqrt{e} \]

Step 8: Integrate with Respect to θ

Next, integrate with respect to \(\theta\): \[ \int_{0}^{2\pi} (2 - \sqrt{e}) \, d\theta \] This evaluates to:\[ (2 - \sqrt{e}) \cdot 2\pi = 4\pi - 2\pi\sqrt{e} \]

Step 9: Final Result

The integral over the region in polar coordinates is: \[ 4\pi - 2\pi\sqrt{e} \]

Key Concepts

Polar Coordinates

Understanding polar coordinates is an essential foundation when dealing with problems involving regions in a circular setting. In polar coordinates, each point on the plane is defined by its distance from the origin, called the radius \(r\), and the angle \(\theta\) it makes with the positive x-axis. This is quite different from the Cartesian system, where points are identified by x and y coordinates.

In simpler terms:

• \(x = r \cos(\theta)\)
• \(y = r \sin(\theta)\)

The formula \(x^2 + y^2 = r^2\) succinctly describes the relationship, encapsulating the whole plane with varying angles \(\theta\) and radii \(r\). This method simplifies dealing with circular and other radial symmetric areas.

Double Integral

The double integral is a mathematical tool used to evaluate functions over a two-dimensional region. It is denoted as \(\int \int\), and it allows us to find volumes, areas, and other quantities when the region of integration is complex.

In this problem, a double integral is needed to integrate the function \(f(x, y) = \left[ \ln(x^2 + y^2) \right] / \sqrt{x^2 + y^2}\) over the defined circular region. By converting where the limits of x and y vary into polar coordinates, you achieve a more manageable form, which simplifies solving such integrals.

This clever transformation allows you to utilize the symmetry of the problem, often reducing the difficulty significantly and enabling a straightforward integration process.

Coordinate Transformation

Coordinate transformation is quite powerful, especially in integral calculus, as it simplifies complex two-dimensional integrals. Here, a transformation from Cartesian to polar coordinates was required. Such a transformation reinterprets the variables in terms of angles and radii, which can simplify expression and analysis of radial functions.

The key transformations:

• \(x = r \cos(\theta)\)
• \(y = r \sin(\theta)\)
• The variable \(\sqrt{x^2 + y^2}\) simplifies to \(r\).

These steps usually adjust the integral into a form that's more tractable and reveal properties or symmetries we had not seen in the Cartesian form.

This is particularly advantageous when integrating over circular regions, like in the exercise.

Area Element in Polar Coordinates

The change to polar coordinates necessitates modifying the area element from the Cartesian form \(dx \, dy\) to \(r \, dr \, d\theta\). This change is crucial for accurately representing the area in terms of the polar parameters.

Why do we do this? The region we are integrating over is defined in terms of circles — concentric and encompassing different \(\theta\) — so the area element in polar coordinates accounts for these circumstances better.

This modification helps us calculate areas correctly:

• \(r \, dr\) represents an infinitesimally thin radial strip.
• \(d\theta\) accounts for the angular expansion of the strip.

Thus, when you set up the double integral in polar form, this transformation tailors the measure of area to fit the region of interest, ensuring the integration is correctly computed.